Question:
If $A$ satisfies the equation $x^{3}-5 x^{2}+4 x+\lambda=0$ then $A^{-1}$ exists if
(a) $\lambda \neq 1$
(b) $\lambda \neq 2$
(c) $\lambda \neq-1$
(d) $\lambda \neq 0$
Solution:
(d) $\lambda \neq 0$
$A$ satisfies $x^{3}-5 x^{2}+4 x+\lambda=0$
$\Rightarrow A^{3}-5 A^{2}+4 A=-\lambda$
Assuming $A^{-1}$ exists, we get
$A^{-1}\left(A^{3}-5 A^{2}+4 A\right)=-\lambda A^{-1}$
$\Rightarrow A^{2}-5 A+4=-A^{-1} \lambda$
$\Rightarrow A^{-1}=\frac{-\left(A^{2}-5 A+4\right)}{\lambda}$
Thus, $\mathrm{A}^{-1}$ exists if $\lambda \neq 0$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.